Four liters of water are equilibrated with a gas mixture containing carbon dioxi

Question

                    Four liters of water are equilibrated with a gas mixture containing carbon dioxide at a partial pressure of .3 atm. Henry's law constant for H2CO3* solubility                     is 2.0 g/L-atm.                 

                    a. How many grams of Co2 are dissolved in the water? What is the pH of this water? (Neglect the secondary dissociation of carbonic acid.)                 

                    b. Assume that H2CO3* dissolves first without dissociating, achieving equilibrium with the atmosphere, and then the solution is closed to the atmosphere.                     H2CO3* then dissociates (neglect the secondary dissociation). Find the PH of this solution and compare it to the result obtained in (a).                 

Explanation / Answer

p=kh*c

P=0.3atm

Kh = 62/2.0 atm/M = 31.0atm/M [given constant has units of g/l*atm. Actual constant has unitsL*atm/mol]

[Mol. wt. of carbonic acid = 62g/mol]

Concentration "C" of H2CO3 in water-gas equilibriated mixture = 0.3/31 M = 0.0097 M

Moles of CO2 dissolved = 0.0097*4 =0.039 M

Weight of CO2 dissolved = 0.039*44=1.703g

now, H2CO3 dissociates.

H2CO3<=> H+ + HCO3-

0.0097-x x x

ka = x^2/(0.0097-x)

2.5*10^(-4) = x^2/(0.0097-x)

[H+] = 1.437*10^(-3) M

pH = -log[H+] = 2.842

b) 0.04% is the concentration of CO2 in atmosphere

Partial pressure of CO2 = 0.0004atm

.Concentration of H2CO3 = 0.0004/31 = 1.3*10^(-5) M

H2CO3 <=> H+ + CO3-

1.3*10^(-5)-x x x

2.5*10^(-4) = x^2/(1.3*10^(-5)-x)

x= 1.2386*10^(-5)

pH = -log[H+] = 4.907

Since, the partial pressure of CO2 is lower in atmosphere, the pH is comparatively higher than the gas mixture used for aeration.

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